Suppose the witch weighs W pounds, and the duck weighs D pounds. Also, suppose their average weight is A pounds. (Their combined weight is 2A)

W + D = 2A. (Given)
W = 2A - D. (Subtract D)
W - 2A = -D (Subtract 2A)
W(W - 2A) = W(-D) (Multiply by W)
W(W - 2A) = (2A - D)(-D) (Substitute 2A-D for W using the given equation)
W² - 2AW = -2AD + D² (Distribute)
W² - 2AW + A² = D² - 2AD + A² (Add A²)
(W-A)(W-A)=(D-A)(D-A). (Factor)
(W-A)² = (D-A)². (An expression times itself equals the expression squared)
(W-A) = (D-A) (Square root)
W = D (Add A)

In other words, a witch weighs the same as a duck.

Where did I go wrong?

Lets say the duck weighs 5 pounds and the witch weighs 15 pounds the average is 10 pounds so if you put this into the equation above

W + D = 2A. (Given)  15+5=2(10)  20=20

so you place the numbers in each time and solve using those numbers.

15 = 2(10)- 5. (Subtract D) 15=15
15 - 2(10) = -5 (Subtract 2A) -5=-5
15(15 - 2(10) = 15(-5) (Multiply by W) -75=-75
15(15 - 2(10)) = (2(10) - 5)(-5) (Substitute 2A-D for W using the given equation) -75=-75
15² - 2(10)(15) = -2(10)(5) + 5² (Distribute) -75=125 This is where the equation fails and is no longer equal to one another
15² - 2(10)(15) + 10² = 5² - 2(10)(5) + 10² (Add A²)
(15-10)(15-10)=(5-10)(5-10). (Factor)
(15-10)² = (5-10)². (An expression times itself equals the expression squared)
(15-10) = (5-10) (Square root) 
15 = 5 (Add A)

the mistake is not with the math it is with the question, if you have the duck and witches weight equal to twice the average weight, then you would need to do a system of equations saying that d=whatever the letter for his weight is ex. x and w=whatever the letter for the witches weight would be ex. y and then (x+y)/2=A and possibly and then w+d=x+y there are other equations you could add also that would be true and would give you an accurate ratio of the weights.

Posted by Robyn on 2005-02-24 01:10:08